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LFM Pure
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Problem Text
Give a careful argument to show that, if $G_{1}$ and $G_{2}$ are subgroups of a finite group $G$ such that every element of $G$ is either in $G_{1}$ or in $G_{2},$ then either $G_{1}=G$ or $G_{2}=G$. Give an example of a group $H$ which has three subgroups $H_{1},H_{2}$ and $H_{3}$ such that every element of $H$ is either in $H_{1},H_{2}$ or $H_{3}$ and $H_{1}\neq H,H_{2}\neq H,H_{3}\neq H$.
Solution (Optional)
Suppose $|G_1|, |G_2| < |G|$ for sake of contraction. Then by Lagrange's theorem $|G_1| \mid |G|$ and $|G_2| \mid |G|$, so $|G_1|, |G_2| \leq \frac{|G|}{2}$. But $|G_1 \cup G_2| = |G_1| + |G_2| - |G_1 \cap G_2|$. $|G_1 \cup G_2| = |G|$ by assumption, and $e \in G_1 \cap G_2$, so $|G_1 \cap G_2| \geq 1$. Therefore $|G| = |G_1| + |G_2| - |G_1 \cap G_2| \leq \frac{|G|}{2} + \frac{|G|}{2} - 1 = |G| - 1 < |G|$, contradiction! Let $H = K_4 = \{e, a, b, c\}$ with $a^2 = b^2 = c^2 = e$. then $H = \{e, a\} \cup \{e, b\} \cup \{e, c\}$ with all subgroups distinct
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